Question

A conical water tank with vertex down has a radius of 11 feet at the top and is 27 feet high. If water flows into the tank at a rate of 10 ft^(3)/min, how fast is the depth of the water increasing when the water is 12 feet deep?

Answer 1

think about the variables given...

you're given the rate of change of volume, and asked for the rate of change of the height

so you need something relating v and h

v = 1/3 * pi * r^2 * h

now you need to express r in terms of h. use similar triangles.

h/27 = r/11 so r = 11h/27

v = 1/3 * pi * (11h/27)^2 * h = pi * 121h^3/(3*27*27)

so dv/dt = 3*pi*121*h^2/(3*27*27) * dh/dt

dv/dt = pi*121*h^2/(27)^2 * dh/dt

if dv/dt = 10, then when h = 12:

dh/dt = 10*(27)^2/(pi*121*144)

Answer 2

thank you so much